Q14 (b) (iv)

Criteria	Marks
Provides correct solution	2
Uses part (iii) to write $J_{k+1}$ in terms of $J_k$, or equivalent merit	1

Sample answer:
$J_n=-\frac1e+nJ_{n-1}\quad n\ge0$
For $n=0$,

$$\text{RHS}=0!-\frac{0!}{e}\sum_{r=0}^{0}\frac1{r!}=1-\frac1e=J_0$$

$=\text{LHS}$, so the property is true for $n=0$.

Assume the property holds for $k$, so
$J_k=k!-\frac{k!}{e}\sum_{r=0}^{k}\frac1{r!}$
By part (iii),

$$\begin{aligned} J_{k+1}&=(k+1)J_k-\frac1e\\ &=(k+1)\left(k!-\frac{k!}{e}\sum_{r=0}^{k}\frac1{r!}\right)-\frac1e\\ &=(k+1)!-\frac{(k+1)!}{e}\left(\sum_{r=0}^{k}\frac1{r!}+\frac1{(k+1)!}\right)\\ &=(k+1)!-\frac{(k+1)!}{e}\sum_{r=0}^{k+1}\frac1{r!} \end{aligned}$$

Thus, the result is true for $n=k+1$, if true for $n=k$. Thus, using the principle of mathematical induction, the result is proven.